Trying to subtract 2 fractional

Question

I'm trying to solve $f(x)=0$ for $x$, but I can't figure it out as I have to get both the denominators to become for instance $8x$, and then only 1 numerator has $x$ in it. How can I solve this?

$$ f(x) = \frac{1}{4}-\frac{3}{2x} $$

Answer 1

$\frac{1}{4} = \frac{3}{2x}$ would be my suggestion and then multiply both sides by "$4x$" to remove the denominators giving:

$x = 6$

Then, one could evaluate the original expression where $f(6) = \frac{1}{4}-\frac{3}{2×6} = \frac{1}{4}-\frac{3}{12} = 0$ which is the desired result.

Answer 2

We have

$$\frac{1}{4}+\frac{-3}{2x}\,,\,\,\,x\neq 0.$$

You can use the product of the denominators as a common denominator... how to do this though?

One way is to take advantage of the following axioms and theorem:

1. $x\cdot 1=x$ for all $x\in\mathbb{R}$.
2. $\displaystyle \frac{x}{x}=1$ for all $x\neq 0$.
3. $\displaystyle \frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd},$ for $b,\,d\neq 0$ but otherwise real.

Now carefully multiply one of the terms by $\displaystyle 1=\frac{2x}{2x}$ and the other by $\displaystyle 1=\frac{4}{4}$ in order to get the job done...

Answer 3

How it can be done one component at a time is shown on mouse rollover:

$$f(x) = \frac14 - \frac{3}{2x} = 0$$

$\frac14-\frac{3}{2x}+\frac{3}{2x}=0+ \frac{3}{2x}$

$$\frac14 = \frac3{2x}$$

$\frac{1}{4} = \frac3{2x}*\frac22$

$$\frac14=\frac{6}{4x}$$

$\frac14 * 4=\frac{6}{4x} * 4$

$$1=\frac6x$$

$1*x = \frac6x * x$

$$x=6$$